If gravity is reality, i CHALLENGE YOU to show the world how one object will orbit another object using gravity, if gravity is reality then this should be simple and easy to show with a science experiment.. once gravity is proven in reality we can end the debate

Show as in math or show as in a physical demonstration?

If math, it is quite simple, assuming there is only one significant body (i.e. one body of a significant mass)

The acceleration due to gravity is given by a=GM/r^2.

The acceleration required to orbit in a circle around an object is given by a=v^2/r.

To orbit, these need to be equal, i.e.:

GM/r^2=v^2/r

Thus GM/r=v^2.

Plug in the numbers and you get an orbit.

Showing it physically, there are countless satellites in orbit around Earth.

If you mean a small model then you need to go outside Earth's Roche limit to even stand a change.

You can't have it supported anywhere near Earth, it needs to be in free fall.

To explain why:

Lets say you have Earth of mass M, and a smaller object of mass m, a distance of R away from the centre of Earth, with an even smaller object trying to orbit it at a distance of r. (capitals for Earth, lower case for smaller object).

Also, lets assume that R>>r, and thus R-r~=r~=R+r.

Thus the acceleration from Earth's gravity on the tiny object will be:

a=GM/R^2 (technically GM/(R-r)^2 at closest approach, or with R+r at furthest)

From the smaller object, it will be:

a=Gm/r^2.

In order to stand a chance at orbiting, M/R^2 needs to be much smaller than m/r^2

Putting in M=6E+24 kg and R= 6.4E+6 m, and m=1000 kg (as an example, we can discuss other masses later) then for these to just be equal you need:

M/R^2=m/r^2, thus r^2=mR^2/M=8.26E-5 m, i.e. 82 um.

I am yet to see an object with a mass of 1000 kg which fits into a sphere of radius 80 microns, and it can't orbit inside another object.

Even if it would work there, you wouldn't easily be able to see the orbit.

We can consider other masses and sizes by considering orbiting just above the surface, so r is the radius of the object.

Then m=4/3*pi*r^3*p, where p is the density.

This then gives (for the acceleration from Earth and the object to be equal):

M/R^2=(4/3*pi*r^3*p)/r^2=4/3*pi*r*p

And thus r=M*3/(R^2*4*pi*p)

The most dense object is only around 20 g/ml, or 20 000 kg/m^3.

Thus, using that, we end up with r=~1700 km, comparable to the moon.

But that would already take you to space, and be basically impossible to support.

So you can't have it supported above Earth.

So your next option is in free fall.

Now the difference in R-r and R is important.

Again, lets consider the object just orbiting on the surface.

(The G constant repeatedly appears, so I am going to ignore it for now, it wont change the results)

Now, we have a due to the object of 4/3*pi*r*p still.

But now from Earth it is more complex.

We have the acceleration of the object of am= M/R^2

We have the acceleration of the orbiting object ao=M/(R-r)^2.

So the relative acceleration (or the tidal acceleration) is at=M/(R-r)^2-M/R^2

=R^2*M/(R^2*(R-r)^2)-(R-r)^2*M/(R^2*(R-r)^2)

=M*(R^2-(R-r)^2)/(R^2*(R-r)^2)

=M*(R^2-R^2+2*R*r-r^2)/(R^2*(R-r)^2)

=M*(2*R*r-r^2)/(R^2*(R-r)^2)

=M*(r*(2*R-r))/(R^2*(R-r)^2)

This now looks big and complicated and ugly, but now we can make the approximation again (as we don't end up with a 0 answer), i.e. R-r~=R and 2*R-r~=2*R

This gives:

at=M*(r*2*R)/(R^2*R^2)

=2*M*r/R^3

So to not have the tidal forces tear the object apart (still not even orbitting it), we need this to be less than m/r^2=4/3*pi*r*p.

Again, for equality:

2*M*r/R^3=4/3*pi*r*p

M/R^3=2/3*pi*p

R^3=3*M/(2*pi*p).

R=cubrt(3*M/(2*pi*p))

Again, using osmium for a best case scenario, we end up with R=~5000 km.

So once again, you need to go into space.

So, assuming gravity is true, you cannot have an object orbit another without going to space, at which point you may as well just orbit Earth as a satellite.